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Question

1 gm of water at a pressure of 1.01×105 Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is

A
167cal
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B
500cal
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C
540cal
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D
581cal
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Solution

The correct option is B 500cal
dW=PΔV = 1.01×105[16711]×106 joule
= 1.01×1674.2cal
= 40 cal. nearly
ΔQ=mL=1×540,
ΔQ=ΔW+ΔU
or ΔU =54040
=500cal.

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